The distance from the center of the Earth to the center of the Moon is called the Lunar Distance. At its closest, this distance is 356,500\ km. Subtracting the radius of the Earth \( r_{Earth} = 6.378137*10^6\ meters \), we have the distance from the surface of the Earth to the center of the Moon:

\( 356500\ km\ -\ r_{Earth} = 3.501*10^8 meters\)

Since Luna’s eyes are closer than the moon, they should appear bigger. Apparent size is measured by angular diameter \( \delta = 2\arctan(\frac{d}{2*D})\), where d is the diameter of the object, and D is the distance to the object. Finding \( D_\Lambda \), the distance to Luna’s eye, requires choosing a point of perspective.

From any point on Earth, the view to Luna’s eyes is blocked by the Earth, Luna’s toes, or 140 Earth volumes of giant girl. Let’s pretend that isn’t true for Endymion, the guy from Gravity. We can reuse some calculations from Gravity too: the central angle between him and Luna is \(\Delta\sigma = 0.20970853\ radians\). We already know the height of Luna’s eyes off the surface of the Earth \( R = 2.20*10^8\ meters \) from earlier. Knowing those two numbers and the radius of the Earth, we can use the law of cosines to calculate \( D_\Lambda \).

Just like in Gravity, the number looks the same with significant figures but is worth calculating. Now we can calculate angular diameter.

\( \delta_\Lambda = 2\arctan(\frac{d}{2*D_\Lambda}) \)

\( = 1.58\ radians \)

In a similar way, we can use the maximum lunar distance (subtracting Earth radius) \( D_{max} = 400329\ km \) to find the maximum angular diameter of the moon \( \delta_{max} = .00868\ radians \) and the minimum lunar distance \(D_{max} = 350129\ km\) to find minimum angular diameter \( \delta_{max} = .00993\ radians \). So we know that Luna’s eyes can appear to be anywhere from 1.592 to 1.821 times the size of the moon depending on the time of month.